(a) Direction ratios of the normal of the planes $7 x+5 y+6 z+30=0$ are $7,5,6$ Direction ratios of the normal of the plane $3 \mathrm{x}-\mathrm{y}-10 \mathrm{z}+4=0$ are $3,-1,-10$ The plane $7 x+5 y+6 z+30=0$ $3 x-y-10 z+y=0$ are perpendicualr to each other if
$$
\begin{aligned}
&\mathrm{a}_1 \cdot \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2=0 \\
&\text { or } 7 \cdot 3+5 \cdot(-1)+6(-10)=21-5-60 \neq 0
\end{aligned}
$$
$\therefore \quad$ Planes (i) and (ii) are not perpendicualr Direction ratio of normal of the plane (i) and (ii) are not proportional as $\frac{7}{3} \neq \frac{5}{-1} \neq \frac{6}{4}$
$\therefore \quad$ These planes are not parallel. Angle between the planes is given by
$$
\cos \theta=\frac{\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2}{\sqrt{\mathrm{a}_1^2+\mathrm{a}_2^2+\mathrm{a}_3^2} \sqrt{\mathrm{b}_1^2+\mathrm{b}_2^2+\mathrm{b}_3^2}}=\frac{-2}{5}
$$
$$
\therefore \quad \theta=\cos ^{-1}\left(-\frac{2}{5}\right) \text {. }
$$
(b) $2 x+y+3 z-2=0$ and $x-2 y+5=0$
Now $\vec{p}_1=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{p}_2=\hat{i}-2 \hat{j}$ $\vec{p}_1 \cdot \vec{p}_2=0 \quad \therefore \vec{p}_1 \perp \vec{p}_2$
hence planes are perpendicular.
(c) $2 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}+5=0$ and $3 \mathrm{x}-3 \mathrm{y}+6 \mathrm{z}-1=0$
Now $\vec{p}_1=2 \hat{i}-2 \hat{j}+4 \hat{k}$ and $\vec{p}_2=3 \hat{i}-3 \hat{j}+6 \hat{k}$
Now $\frac{a_1}{a_2}=\frac{2}{3}, \frac{b_1}{b_2}=\frac{2}{3}, \frac{c_1}{c_2}=\frac{2}{3}$
$\therefore \quad \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \Rightarrow$ Normals are parallel.
Hence, planes are parallel.
(d) $2 x-y+3 z-1=0$ and $2 x-y+3 z+3=0$
Now
$\vec{p}_1=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{p}_2=2 \hat{i}-\hat{j}+3 \hat{k}$
Now $\frac{a_1}{a_2}=\frac{2}{2}=1, \frac{b_1}{b_2}=\frac{-1}{-1}=1, \frac{c_1}{c_2}=\frac{3}{3}=1$
$$
\therefore \quad \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}
$$
$\therefore$ Normals are parallels. Hence planes are parallels.
(e) $4 x+8 y+z-8=0$ and $y+z-4=0$
$\vec{p}_1=4 \hat{i}+8 \hat{j}+\hat{k}$ and $\quad \vec{p}_2=\hat{j}+\hat{k}$
$\vec{p}_1 \cdot \vec{p}_2=4(0)+8(1)+1(1)=9$
$$
\begin{aligned}
&\left|\vec{p}_1\right|=\sqrt{4^2+8^2+1^2}=\sqrt{81} \text { and } \\
&\left|\vec{p}_2\right|=\sqrt{1^2+1^2}=\sqrt{2}
\end{aligned}
$$
Angle between planes is the angle between their normals. Let $\theta$ be the angle between the planes.
$$
\therefore \quad \cos \theta=\frac{\vec{p}_1 \cdot \vec{p}_2}{\left|\vec{p}_1 \| \vec{p}_2\right|}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ} .
$$